Now $a c-b^2=2$ and $2 a+b=1$ and $2 b + c =2$
solving all these above equations we get
$\frac{1-b}{2} \times\left(\frac{2-2 b}{1}\right)-b^2=2$
$\Rightarrow(1-b)^2-b^2=2$
$\Rightarrow 1-2 b=2$
$\Rightarrow b=-\frac{1}{2} \text { and } a=\frac{3}{4} \text { and } c=3$
Hence $\alpha=3 a+\frac{3 b}{2}=\frac{9}{4}-\frac{3}{4}=\frac{3}{2}$
and $\beta=3 b+\frac{3 c}{2}=-\frac{3}{2}+\frac{9}{2}=3$
also $s = a + c =\frac{15}{4}$
$\therefore \frac{\beta s}{\alpha^2}=\frac{3 \times 15}{4 \times \frac{9}{4}}=5$
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